What is the arc length of #f(t)=(3te^t,t-e^t) # over #t in [2,4]#?

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Answer 1 · 2017-05-15T12:41:30

# 612.530 # (3dp)

We have:

# f(t) = (3te^t, t-e^t ) # where #t in [2,4]#

The parametric arc-length is given by:

# L = int_(alpha)^(beta) \ sqrt((dx/dt)^2 + (dy/dt)^2 ) \ dt #

We can differentiate the parameters:

# x(t) = 3te^t => dx/dt = 3te^t + 3e^t #

# y(t) = t-e^t => dy/dt = 1-e^t #

Then the arc-length is given by:

# L = int_2^4 \ sqrt( (3te^t + 3e^t)^2 + (1-e^t)^2 ) \ dt #

This integral dos not have a trivial anti-derivative, and so is evacuated using numerical methods to give:

# L = 612.530 # (3dp)